MHT CET 2023 12th May Morning Shift | Probability Question 21 | Mathematics

$$\mathrm{A}$$ and $$\mathrm{B}$$ are independent events with $$\mathrm{P}(\mathrm{A})=\frac{1}{4}$$ and $$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$$, then $$\mathrm{P}(\mathrm{B})$$ is

$$\frac{1}{4}$$

$$\frac{3}{5}$$

$$\frac{2}{3}$$

$$\frac{2}{5}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Then mean of number of tens is

$$\frac{1}{13}$$

$$\frac{1}{169}$$

$$\frac{2}{13}$$

$$\frac{4}{169}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

A fair die is tossed twice in succession. If $$\mathrm{X}$$ denotes the number of fours in two tosses, then the probability distribution of $$\mathrm{X}$$ is given by

$$\mathrm{X}=x_i$$	0	1	2
$$\mathrm{P_i}$$	$$\frac{1}{36}$$	$$\frac{25}{36}$$	$$\frac{5}{18}$$

$$\mathrm{X}=x_i$$	0	1	2
$$\mathrm{P_i}$$	$$\frac{25}{36}$$	$$\frac{1}{36}$$	$$\frac{5}{18}$$

$$\mathrm{X}=x_i$$	0	1	2
$$\mathrm{P_i}$$	$$\frac{25}{36}$$	$$\frac{5}{18}$$	$$\frac{1}{36}$$

$$\mathrm{X}=x_i$$	0	1	2
$$\mathrm{P_i}$$	$$\frac{5}{18}$$	$$\frac{1}{36}$$	$$\frac{25}{36}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

If $$\mathrm{A}$$ and $$\mathrm{B}$$ are two events such that $$\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{3}$$, then the value of $$\mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right)$$ is

$$\frac{5}{6}$$

$$\frac{1}{6}$$

$$\frac{11}{6}$$

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